Question: $f(3)=4\,$, $~f\,^\prime(3)=5\,$, $~f\,^{\prime\prime}(3)=\frac{1}{4}\,$, and $~f\,^{\prime\prime\prime}(3)=\frac{2}{3}\,$. What are the first four nonzero terms of the Taylor series, centered at $x=3$, of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $4+5x+\frac{1}{8}{{x}^{2}}+\frac{1}{9}{x^{3}}$ (Choice B) B $4+5x+\frac{1}{4}{{x}^{2}}+\frac{2}{3}{x^{3}}$ (Choice C) C $4+5(x+3)+\frac{1}{8}{{(x+3)}^{2}}+\frac{1}{9}{{(x+3)}^{3}}$ (Choice D) D $4+5(x-3)+\frac{1}{8}{{(x-3)}^{2}}+\frac{1}{9}{{(x-3)}^{3}}$ (Choice E) E $4+5(x-3)+\frac{1}{4}{{(x-3)}^{2}}+\frac{2}{3}{{(x-3)}^{3}}$
Solution: We know the formula for a Taylor series centered at $~x=3~$ for the function $~f~$ is $ f(3)+f\,^\prime(3)(x-3)+\frac{f\,^{\prime\prime}(3)}{2!}{{(x-3)}^{2}}+...+\frac{{{f}^{(n)}}(3)}{n!}{{(x-3)}^{n}}+...\,$ If we substitute the given function and derivative values in the first four terms, we get the following: $ T_3(x)=4+5(x-3)+\frac{\frac{1}{4}}{2!}{{(x-3)}^{2}}+\frac{\frac{2}{3}}{3!}{{x}^{3}}\,$. This simplifies to $ T_3(x)=4+5(x-3)+\frac{1}{8}{{(x-3)}^{2}}+\frac{1}{9}{{(x-3)}^{3}}$.